charlenecsn
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Express 4 sin x + 3 cos x in the form r sin (x + α). Hence find all the values of x in the range 0 ≤ x ≤ 360° for which cos 3x = cos 2x.

I‘ve done the first part, which I got 5 sin (x + 36.9°), but I have no idea how to solve the second part.

Can anyone help please, thanks!
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mqb2766
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Can you upload a picture of the question?
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charlenecsn
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Name:  B31495CE-6624-4B6D-A1B1-29EE22521F01.jpg.jpeg
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mqb2766
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I think they want you to.expand the multiple angles in terms of cos(x) and sin(x).
However, id just take acos directly and forget about the first part.
Last edited by mqb2766; 4 weeks ago
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charlenecsn
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(Original post by mqb2766)
I think they want you to.expand the multiple angles in terms of cos(x) and sin(x).
However, id just take acos directly and forget about the first part.
I‘ve tried to expand it in terms of cosx and sinx but I can’t link back to the first equation.
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CaptainDuckie
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Cos(A+_B) = CosAcosB + SinASinB
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mqb2766
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(Original post by charlenecsn)
I‘ve tried to expand it in terms of cosx and sinx but I can’t link back to the first equation.
Can you upload?
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charlenecsn
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(Original post by CaptainDuckie)
Cos(A+_B) = CosAcosB + SinASinB
I got 4cos^3(x)-3cos(x)=2cos^2x-1, but I can't relate this to the first equation.
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CaptainDuckie
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Upload pic bro
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charlenecsn
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(Original post by mqb2766)
Can you upload?
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CaptainDuckie
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Cannot see that at all😂😂
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charlenecsn
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(Original post by CaptainDuckie)
Cannot see that at all😂😂
you mean the pic cant be viewed?😂
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CaptainDuckie
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(Original post by charlenecsn)
you mean the pic cant be viewed?😂
Your writing
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charlenecsn
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(Original post by CaptainDuckie)
Your writing
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mqb2766
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(Original post by charlenecsn)
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Will have a look in the morning, but the simple thing to do is take acos
to give
3x = +/-2x + 2*pi*n
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charlenecsn
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(Original post by mqb2766)
Will have a look in the morning, but the simple thing to do is take acos
to give
3x = +/-2x + 2*pi*n
Alright, thank you so much!!!
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simon0
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As a side note, another method is to use the trigonometric sum-product results, specifically:

 \cos (a-b) - \cos (a+b) = 2 \sin(a) \sin(b) .

For this case:  \cos (2x) - \cos (3x) = 2 \sin ( (5/2)x ) \sin( (3/2) x) = 0 .
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charlenecsn
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(Original post by simon0)
As a side note, another method is to use the trigonometric sum-product results, specifically:

 \cos (a-b) - \cos (a+b) = 2 \sin(a) \sin(b) .

For this case:  \cos (2x) - \cos (3x) = 2 \sin ( (5/2)x ) \sin( (3/2) x) = 0 .
Thanks! ))but that formula isn't in the syllabus
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simon0
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(Original post by CaptainDuckie)
Cos(A+_B) = CosAcosB + SinASinB
Close but not quite.
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CaptainDuckie
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(Original post by simon0)
Close but not quite.
It’s the Sin(A _B) = SinASinB - CosAcosB

( I realised )

But forgot to edit
Last edited by CaptainDuckie; 4 weeks ago
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